How does one solve this kind of equation: $3^x=x+3$

Question

How does one solve this kind of equation: $$3^x=x+3$$

I tried playing around with logs but it didn't get me anywhere. I plotted the two functions $f(x)=3^x$ and $g(x)=x+3$ on a graph to estimate the point of intersection but how do I solve it?

Answer 1

You've to use the Product log function (when you don't want it the 'numerical method' way):

$$y^x=x+y\Longleftrightarrow x=-\frac{\text{W}_n\left(-\frac{\ln(y)}{y^y}\right)+y\ln(y)}{\ln(y)}$$

Where $\text{W}_k(z)$ is the analytic continuation of the product log funtion and $n\in\mathbb{Z}$.

When $y=3$, we've the following real solutions:

• $$x=-\frac{\text{W}\left(-\frac{\ln(3)}{27}\right)+3\ln(3)}{\ln(3)}\approx-2.96136$$
• $$x=-\frac{\text{W}_{-1}\left(-\frac{\ln(3)}{27}\right)+3\ln(3)}{\ln(3)}\approx1.33509$$

Answer 2

Here is my solution using the Newton-Raphson method.

$$f(x)=3^x-x-3$$

$$f'(x)=3^xln3-1$$

$$x_{n+1}=x_{n}-\frac{f(x)}{f'(x)}$$

Getting first estimates from my graph: $$Let: x_{1}=2$$ $$x_{2}=1.5499$$ $$x_{3}=1.3632$$ $$x_{4}=1.3356$$ $$x_{5}=1.3351$$ $$x=1.3 (1dp)$$

And then the second point of intersection: $$Let: x_{1}=-1$$ $$x_{2}=-3.630$$ $$x_{3}=-2.968$$ $$x_{4}=-2.961$$ $$x=-3.0 (1dp)$$

Thanks all.